Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 1e5 + 100;
int n, m, q;
int a[N];
std::vector<int> edge[N];
std::vector<int> set[N][23];
int father[N][23];
int deep[N];
void init() {
  std::cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int u;
    scanf("%d", &u);
    set[u][0].push_back(i);
  }
}
void dfs(int u) {
  for (int i = 0; i < (int)edge[u].size(); i++) {
    int v = edge[u][i];
    if (v == father[u][0]) {
      continue;
    }
    father[v][0] = u;
    deep[v] = deep[u] + 1;
    dfs(v);
  }
}
std::vector<int> merge(std::vector<int> s1, std::vector<int> s2) {
  std::vector<int> ret;
  for (int i = 0; i < (int)s1.size(); i++) {
    ret.push_back(s1[i]);
  }
  for (int i = 0; i < (int)s2.size(); i++) {
    ret.push_back(s2[i]);
  }
  std::sort(ret.begin(), ret.end());
  while ((int)ret.size() > 10) {
    ret.pop_back();
  }
  return ret;
}
void prepare() {
  for (int i = 1; i <= n; i++) {
    std::sort(set[i][0].begin(), set[i][0].end());
    while ((int)set[i][0].size() > 10) {
      set[i][0].pop_back();
    }
  }
  for (int j = 1; j <= 20; j++) {
    for (int i = 1; i <= n; i++) {
      father[i][j] = father[father[i][j - 1]][j - 1];
      set[i][j] = merge(set[i][j - 1], set[father[i][j - 1]][j - 1]);
    }
  }
}
int leastCommonAncestor(int u, int v) {
  if (deep[u] < deep[v]) {
    std::swap(u, v);
  }
  for (int i = 20; i >= 0; i--) {
    if (deep[father[u][i]] >= deep[v]) {
      u = father[u][i];
    }
  }
  if (u == v) {
    return u;
  }
  for (int i = 20; i >= 0; i--) {
    if (father[u][i] != father[v][i]) {
      u = father[u][i];
      v = father[v][i];
    }
  }
  return father[u][0];
}
void work() {
  deep[1] = 1;
  dfs(1);
  prepare();
  for (int i = 1; i <= q; i++) {
    int u, v, need;
    scanf("%d%d%d", &u, &v, &need);
    int lca = leastCommonAncestor(u, v);
    std::vector<int> vec[2];
    for (int i = 20; i >= 0; i--) {
      if (deep[father[u][i]] >= deep[lca]) {
        vec[0] = merge(vec[0], set[u][i]);
        u = father[u][i];
      }
      if (deep[father[v][i]] >= deep[lca]) {
        vec[1] = merge(vec[1], set[v][i]);
        v = father[v][i];
      }
    }
    std::vector<int> answer = merge(vec[0], vec[1]);
    answer = merge(answer, set[lca][0]);
    std::sort(answer.begin(), answer.end());
    while ((int)answer.size() > need) {
      answer.pop_back();
    }
    printf("%d ", (int)answer.size());
    for (int i = 0; i < (int)answer.size(); i++) {
      printf("%d ", answer[i]);
    }
    printf("\n");
  }
}
int main() {
  init();
  work();
  return 0;
}